What is the integral of (2x+1)/(x^2+2x+17) ?

The integral of (2x+1)/(x^2+2x+17) is ln|x^2+2x+17|-1/4 arctan(1/4 (x+1))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2x+1)/(x^2+2x+17)

\int \frac{2 x + 1}{x ^{2} + 2 x + 17} d x

ln x^{2} + 2 x + 17 - \frac{1}{4} arctan (\frac{1}{4} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 17} d x

Complete the square

x^{2} + 2 x + 17 : (x + 1)^{2} + 16

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 16} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 16} d u

= \int \frac{2 u - 1}{u ^{2} + 16} d u

Expand

\frac{2 u - 1}{u ^{2} + 16} : \frac{2 u}{u ^{2} + 16} - \frac{1}{u ^{2} + 16}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{2 u}{u ^{2} + 16} d u - \int \frac{1}{u ^{2} + 16} d u

\int \frac{2 u}{u ^{2} + 16} d u = ln u^{2} + 16

\int \frac{1}{u ^{2} + 16} d u = \frac{1}{4} arctan (\frac{u}{4})

= ln u^{2} + 16 - \frac{1}{4} arctan (\frac{u}{4})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 16 - \frac{1}{4} arctan (\frac{x + 1}{4})

Simplify

ln (x + 1)^{2} + 16 - \frac{1}{4} arctan (\frac{x + 1}{4}) : ln x^{2} + 2 x + 17 - \frac{1}{4} arctan (\frac{1}{4} (x + 1))

= ln x^{2} + 2 x + 17 - \frac{1}{4} arctan (\frac{1}{4} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 17 - \frac{1}{4} arctan (\frac{1}{4} (x + 1)) + C

\int_{0}^{1} 13 cos (x^{2}) d x

- csc (x)

\int \frac{49 - x ^{2}}{x} d x

(0.4 x + 2) (0.5 x - 5)

\frac{d}{d x} (x^{7} e^{10 x})

What is the integral of (2x+1)/(x^2+2x+17) ?
The integral of (2x+1)/(x^2+2x+17) is ln|x^2+2x+17|-1/4 arctan(1/4 (x+1))+C